Please expand the following at x = 0: sqrt(1 + x)^ -1. Thank you for help.
(1 + x)^ -1/2 at x = 0
Using the Maclaurin Series, we have
f(0) + f ' (0)x + f ''(0) x^2 / 2! + f '''(0) x^3 / 3! + ...... + f(n)(0) x^n / n!
f(0) = 1
f ' (x) = ( -1/2) (1 + x)^( -3/2) and f ' (0) = (-1.2) ( 1 + 0)^-(3/2)= (-1/2)
f '' (x) = (3/4) (1 + x) ^(-5/2) and f '' (0) = (3/4) (1 + 0)^(-5/2) = (3/4)
f ''' (x) = (-15 / 8 ) ( 1 + x)^(-7/2) and f''' (0) = (-15/8) (1 + 0)^(-7/2) = ( - 15 / 8)
So we have the first four terms
1 - (1/2)x + (3/4)x^2/2 - (15/8)x^3 / 6 + ..... + f(n)(0) x^n / n! =
1 - x/2 + 3x^2/8 - 15x^3/48 + .....+ f(n)(0) x^n / n! =
1 - x/2 + 3x^2/8 - 5x^3/16 + ...... + f(n)(0) x^n / n!